I'm having trouble figuring this out. Can anyone explain to me how to find a reasonable upper bound? I think we use Chebychev's inequality but I'm not sure what to plug in where
stallionaire: March 19, 2015, 2:14 a.m.
wrong board my bad
Katherine: March 19, 2015, 1:01 p.m.
You're on the right track thinking that it's Chebychev's inequality! n=100,000 It is implied that a fair coin has p=1/2 heads. The equidistance between 50,000 and both 60,0000/100,000 and 40,000/100,000 must equal 10,0000/100,000 .1=c P(S100,000-1/2)>.1<[1/(4*100,000*.1^2)]